# Question 9.21: (a) Find the inverse of A = (2 1 1 6 5 -3 4 -1 3) by the eli......

(a) Find the inverse of

A = $\begin{pmatrix} 2 & 1 & 1 \\ 6 & 5 & -3 \\ 4 & -1 & 3 \end{pmatrix}$

by the elimination method, hence solve the equations

2x + y + z = 12
6x + 5y − 3z = 6
4x − y + 3z = 5

(b) Solve the equations in (a) by Gauss–Jordan elimination.

Step-by-Step
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(a) Set up the augmented matrix, consisting of the matrix A and the unit matrix of the same dimension. Add multiples of rows to other rows to reduce A to upper triangular form. Enter the Excel formula for each row operation in column 1, then copy the formula across the remaining five columns. See Chapter 1, and remember that formulae start with =.

For example, to divide row 1 by 2, place the cursor in cell B8 and type = B4/2. To copy this formula across, click on the corner of cell B8 until a black cross appears, then drag this across the remaining five columns. This updated row is labelled row 1¹. Similarly, to add row 1¹ × (−6) to row 2 in order to generate the required 0, place the cursor in cell B9 and type in the formula = B5 + B8* (−6). Copy the formula across the row. In this way carry out all the row operations as indicated.

A is reduced to upper triangular form: row 1¹, row 2² and row 3². Continue with the row operations to further reduce A to the unit matrix.

The reduction is now complete. Write out the most recently updated rows in order.

Read off the inverse matrix: columns 4, 5 and 6. The inverse of A is

Finally, to multiply $A^{–1}$ by the column of constants from the RHS of the equations, arrangethese as shown.

(b) To solve by Gauss–Jordan elimination, go back to the augmented matrix in (a) and replace col. 5 (first column of the unit matrix) by the column of constants from the RHS of each equation. Delete col. 5 and col. 6 (see below). To carry out the Gauss–Jordan elimination,

copy across the formulae that were entered earlier to find the inverse of A. Read off the solution.

At this stage it is possible to read off the upper triangular matrix: row 1¹, row 2² and row 3². Solve by back substitution. For Gauss–Jordan elimination continue the row operations.

Read off the solution from the reduced augmented matrix: x=−2.5, y = 9, z = 8.

 1 A B C D E F G 2 Augmented matrix 3 col 1 col 2 col 3 col 4 col 5 col 6 4 row 1 → 2 1 1 1 0 0 5 row 2 → 6 5 -3 0 1 0 6 row 3 → 4 -1 3 0 0 1 7 Operation Formula 8 row 1¹ → 1 0.5 0.5 0.5 0 0 row 1 divided by 2 = B4/2 9 row 2¹ → 0 2 -6 -3 1 0 row 2 + row 1¹ × (-6) = B5 + B8 ∗ (-6) 10 row 3¹ → 0 -3 1 -2 0 1 row 3 + row 1¹ × (-4) = B6 + B8 ∗ (-4) 11 12 row 1¹ → 1 0.5 0.5 0.5 0 0 13 row 2² → 0 1 -3 -1.5 0.5 0 row 2¹ divided by 2 = B12/2 14 row 3² → 0 0 -8 -6.5 1.5 1 row 3¹ + row 2² × (3) = B10 + B13 ∗ (3)

 15 row 3³ → 0 0 1 0.81 -0.2 -0.1 row 3² divided by -8 = B14 / (- 8) 16 row 1² → 1 0.5 0 0.09 0.09 0.06 row 1¹ + row 3³ × (-0.5) =B12 + B15 ∗ (-0.5) 17 row 2³ → 0 1 0 0.94 -0.1 -0.4 row 2² + row 3³ × (3) = B13 + B15 ∗ (3) 18 row 1³ → 1 0 0 -0.4 0.13 0.25 row 1² + row 2³ × (-0.5) = B16 + B17 ∗ (-0.5)

 col 1 col 2 col 3 col 4 col 5 col 6 row 1² → 1 0 0 -0.375 0.0125 0.025 row 2³ → 0 1 0 0.9375 -0.0625 -0.375 row 3³ → 0 0 1 0.8125 -0.8125 -0.125

 -0.375 0.125 0.25 0.9375 -0.0625 -0.375 0.8125 -0.1875 -0.3125

 B C D E 19 Inverse of A RHS 20 -0.375 0.125 0.25 12 21 0.9375 -0.0625 -0.375 6 22 0.8125 -0.1875 -0.125 5 23 24 x = -2.5 = B20 * E20 + C20 * E21 + D20 * E22 Formulae used to calculate x, y and z 25 y = 9 = B21 * E20 + C21 * E21 + D21 * E22 26 z = 8 = B22 * E20 + C22 * E21 + D22 * E22

 Augmented matrix row 1 2 1 1 12 row 2 6 5 -3 6 row 3 4 -1 3 5 row 1¹ 1 0.5 0.5 6 row 1 divided by 2 row 2¹ 0 2 -6 -30 row 2 + row 1¹ × (-6) row 3¹ 0 -3 1 -19 row 3 + row 1¹ × (-4) row 2² 0 1 -3 -15 row 2¹ divided by 2 row 3² 0 0 -8 -64 row 3¹ + row 2² × (3)

 row 3³ 0 0 1 8 row 3² divided by – 8 row 1² 1 0.5 0 2 row 1¹ + row 3³ × (-0.5) row 2³ 0 1 0 9 row 2² + row 3³ × (3) row 1³ 1 0 0 -2.5 row 1² + row 2³ × (-0.5)

 col 1 col 2 col 3 col 4 row 1² → 1 0 0 -2.5 row 2³ → 0 1 0 9 row 3³ → 0 0 1 8

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